C 为什么它总是只返回单个链接列表的一个节点?
我有一个结构C 为什么它总是只返回单个链接列表的一个节点?,c,C,我有一个结构 typedef struct song{ string title; string artist; string album; string genre; float rating; struct song *next; }song_t; 和删除功能 song_t *DeleteSong(song_t *head, string key){ song_t *temp, *temp2, *holder;
typedef struct song{
string title;
string artist;
string album;
string genre;
float rating;
struct song *next;
}song_t;
和删除功能
song_t *DeleteSong(song_t *head, string key){
song_t *temp, *temp2, *holder;
holder = (song_t *)malloc(sizeof(song_t));
temp = head;
while(temp != NULL && strcasecmp(temp->title, key) != 0){
temp = temp->next;
}
if(temp == NULL){
newline;
printf("Song not found.");
newline;
newline;
}
else if(temp == head){
if(temp->next == NULL){
temp->next = NULL;
free(temp);
return NULL;
}
else{
head = temp->next;
}
}
else if(temp->next == NULL){
temp2 = head;
while(temp2->next->next != NULL){
temp2 = temp2->next;
}
holder = temp2->next->next;
temp2->next = NULL;
}
else{
temp2 = head;
while(temp2->next != temp){
temp2 = temp2->next;
}
holder = temp;
temp2->next = temp->next;
free(holder);
}
return head;
}
song_t *RemoveDuplicate(song_t *head){
song_t *temp;
song_t *temp2;
temp = head;
temp2 = temp->next;
while(temp != NULL){
while(temp2 != NULL){
if(strcasecmp(temp->title,temp2->title) == 0 && strcasecmp(temp->artist,temp2->artist) == 0 && strcasecmp(temp->album,temp2->album) == 0 && strcasecmp(temp->genre,temp2->genre) == 0){
if(temp2 == temp->next){
temp = DeleteSong(temp,temp2->title);
}
else{
temp2 = DeleteSong(temp2->next,temp2->title);
}
}
temp2 = temp2->next;
}
temp = temp->next;
}
}
最后是一个删除重复函数
song_t *DeleteSong(song_t *head, string key){
song_t *temp, *temp2, *holder;
holder = (song_t *)malloc(sizeof(song_t));
temp = head;
while(temp != NULL && strcasecmp(temp->title, key) != 0){
temp = temp->next;
}
if(temp == NULL){
newline;
printf("Song not found.");
newline;
newline;
}
else if(temp == head){
if(temp->next == NULL){
temp->next = NULL;
free(temp);
return NULL;
}
else{
head = temp->next;
}
}
else if(temp->next == NULL){
temp2 = head;
while(temp2->next->next != NULL){
temp2 = temp2->next;
}
holder = temp2->next->next;
temp2->next = NULL;
}
else{
temp2 = head;
while(temp2->next != temp){
temp2 = temp2->next;
}
holder = temp;
temp2->next = temp->next;
free(holder);
}
return head;
}
song_t *RemoveDuplicate(song_t *head){
song_t *temp;
song_t *temp2;
temp = head;
temp2 = temp->next;
while(temp != NULL){
while(temp2 != NULL){
if(strcasecmp(temp->title,temp2->title) == 0 && strcasecmp(temp->artist,temp2->artist) == 0 && strcasecmp(temp->album,temp2->album) == 0 && strcasecmp(temp->genre,temp2->genre) == 0){
if(temp2 == temp->next){
temp = DeleteSong(temp,temp2->title);
}
else{
temp2 = DeleteSong(temp2->next,temp2->title);
}
}
temp2 = temp2->next;
}
temp = temp->next;
}
}
但是,每当我在main中包含remove duplicate函数时,
头部=拆下的副本(头部)
结果总是只返回一个结构并删除整个列表。我认为RemovedUpplicate函数有问题,因为我测试了DeleteSong函数,它工作得很好。如果列表中的前两个条目是重复的,那么第一次检查时似乎:
if(temp2 == temp->next){
temp = DeleteSong(temp,temp2->title);
}
它将在列表的最前面评估为true。如果您的列表突然减少到列表的最前面,您可能需要检查接下来会发生什么
我怀疑这个错误实际上可能是在DeleteSong函数中发现的,因为它看起来比我所见过的其他单链接列表要复杂一些
在pseudo-C中,我可能会尝试以下操作:
to_delete = find_node_by_key( key )
return if to_delete == null
current = head
last = null
if to_delete == current
{
head = current->next
to_mem_free = current
}
else do
{
if to_delete == current
{
to_mem_free = current
if ( last != null ) last->next = current->next
break
}
last = current
} while (current = current->next) != null
可以将对要删除的节点的搜索与实际删除结合起来,这样就不必遍历列表两次。您可能还希望尽量避免使用“temp”作为变量名,除非绝对必要,因为它通常会导致混淆
许多工具链包括一个提供单链接列表的库。通过使用这样的库,您可以节省数小时的编码和调试时间,该库还可能提供排序树或哈希列表,从而显著加快搜索,例如歌曲标题
holder = (song_t *)malloc(sizeof(song_t));
/* snip */
/* else if */
holder = temp2->next->next;
/* else */
holder = temp;
temp2->next = temp->next;
free(holder);
分配holder=temp2->next->next时
(顺便说一下,这里总是NULL
),或者holder=temp代码>,您正在丢失对malloc
ed内存的引用。由于您根本没有真正使用holder
,因此修复方法是将其从函数中删除。在第一种情况下,除了分配一个复杂的NULL
,您没有以任何方式使用它,在第二种情况下,删除holder=temp编码>和免费
ingtemp
是正确的方法
还有一些奇怪的事情和错误:
song_t *DeleteSong(song_t *head, string key){
song_t *temp, *temp2, *holder;
holder = (song_t *)malloc(sizeof(song_t)); // as said, remove holder completely
temp = head;
/* Find song with given title */
while(temp != NULL && strcasecmp(temp->title, key) != 0){
temp = temp->next;
}
if(temp == NULL){
newline;
printf("Song not found.");
newline;
newline;
}
/* It's the very first in the list */
else if(temp == head){
if(temp->next == NULL){
/* It's even the only one */
temp->next = NULL; // This runs only if temp->next is already NULL
free(temp); // Also free the members of temp, or you're leaking
return NULL;
}
else{
head = temp->next; // You should now free temp and members, or you're leaking memory
}
}
/* It's the last one in the list, but not the first */
else if(temp->next == NULL){
temp2 = head;
/* Find the penultimate song */
while(temp2->next->next != NULL){
temp2 = temp2->next;
}
holder = temp2->next->next;
temp2->next = NULL; // You should now free temp and members, or you're leaking memory
}
/* Neither first nor last */
else{
temp2 = head;
/* Find song before */
while(temp2->next != temp){
temp2 = temp2->next;
}
holder = temp;
temp2->next = temp->next;
free(holder);
}
return head;
}
但除了泄漏,这是正确的,尽管不必要地复杂
song_t *DeleteSong(song_t *head, string key) {
song_t *prev = NULL, curr = head;
/* Find song and node before that */
while(curr != NULL && strcasecmp(curr->title, key) != 0) {
prev = curr;
curr = curr->next;
}
if (curr == NULL) {
/* Not found */
newline;
printf("Song not found.");
newline;
newline;
} else if (prev == NULL) {
/* It's the very first song in the list
* so let head point to its successor
* and free the song; it doesn't matter
* if it's the last in the list
*/
head = curr->next;
free(curr->title); // Probably, but not if title isn't malloced
free(curr->artist); // Ditto
free(curr->album);
free(curr->genre);
free(curr);
} else {
/* We have a predecessor, let that point
* to the successor and free the song
*/
prev->next = curr->next;
free(curr->title); // See above
free(curr->artist);
free(curr->album);
free(curr->genre);
free(curr);
}
return head;
}
另一方面,您的RemovedUpplicate
功能并没有达到您的目的。除了副本紧跟原始版本和不紧跟原始版本之间的区别(我看不出有什么原因),赋值temp=DeleteSong(temp,temp2->title)代码>响应temp2=DeleteSong(temp2->next,temp2->title)代码>更改什么temp
resptemp2
指向,但不指向列表中相应的前置项所指向的位置。让我们用一点ASCII艺术来说明这个问题:
temp temp2
| |
v v
song1->song2->song3->song4->song5->...
其中song2
和song3
是重复的。现在在temp=DeleteSong(temp,temp2->title)中
,因为这两首歌是重复的,DeleteSong
的head
参数已经匹配,并且在您的DeleteSong
版本中,head=temp->next;回流头代码>是所有要做的事情,所以列表根本没有改变,您只需要
temp temp2
| |
v v
song1->song2->song3->song4->...
指向同一列表节点的两个指针。在释放节点的DeleteSong
版本中,song1。下一个现在将指向释放的d内存,哎哟
如果副本没有立即跟随原始版本,temp2=DeleteSong(temp2->next,temp2->title)
可能找不到具有匹配标题的歌曲,因为搜索是在已知匹配之后开始的,在这种情况下,列表根本不会修改,temp2
只是更改为指向后续歌曲。如果之后有一首歌曲具有匹配的标题,找到的副本仍不会从列表中删除,之后的部分可能会更改,可能导致不健全状态。如果temp2
指向列表中倒数第二个节点,并且最后一个节点具有匹配的标题,则最后一个节点是free
d,但复制的next
指针没有更改,因此现在它指向free
d内存,您有一个悬空指针(这是一个等待发生的segfault)
另一个问题是DeleteSong
和RemoveDuplicate
中的删除标准不同,前者只检查标题,后者还检查艺术家、专辑和流派,因此在后者中使用前一个功能可能会删除不应删除的歌曲(考虑封面版本)
如果要从单链表中删除节点,则需要一个指向前一个节点的指针,以便能够更改该节点指向的内容,否则将创建悬空指针。我在上面给出了一种标准方法,您基本上可以将其复制到内部循环中,但在这里,我们永远不可能删除第一个节点,因此更简单一些:
// void, since head is never changed, only duplicates after the original are removed
void RemoveDuplicates(song_t *head) {
song_t *orig = head, prev, curr;
while(orig != NULL && orig->next != NULL) {
prev = orig;
curr = orig->next;
while(curr != NULL) {
// Find next song to remove
while(curr != NULL && !meets_deletion_criteria(orig, curr)) {
prev = curr;
curr = curr->next;
}
// Now either curr is NULL, or it shall be deleted
if (curr != NULL) {
// Let the predecessor point to curr's successor
prev->next = curr->next;
clean_up(curr); // free all malloced members and the node itself
curr = prev->next;
}
}
orig = orig->next;
}
}
在else语句中,我不明白为什么要将temp2->next作为头指针传递。在temp=temp->next
(在内部while之后)之后,您不应该放置temp2=temp->next代码>?